Question

The area of a field shaped like a right triangle is 600 m2.The legs of the
field are fenced with steel at $10/m, while the hypotenuse is fenced with
aluminum at $20/m.The perimeter of the field is 120 m.The total cost
of the fencing is $1700.
a. What is the length of each side fenced with steel?
b. What is the length of the side fenced with aluminum?

Answer 1

U have 600 = xy/2

x+y+h = 120

h^2 = x^2 + y^2

3 equations, 3 unknowns, solve.

Answer 2

Where did you get your middle equation? I had 10x+10y+20h=1700.

Answer 3

Oh i see they give you the perimeter too. Hell, 4 equations, 3 unknowns :)

Answer 4

I am just calling the sides x,y and h, the cost of the sides can be dealt with after u have the sizes.

Answer 5

im still confused as to what im supposed to do

Answer 6

They give you the area. The area of a right triangle is (1/2) base times height. we'll call those x and y respectively. So, A=(1/2)xy. They give you area, so your first equation is 1200=xy. Your second equation is that they give you the perimeter, 120m. so, x+y+h=120. We can stay away from the square roots by using the cost, or 10x+10y+20h=1700.

Answer 7

U may do it by any method u find most easy for u, no?

Answer 8

h=120-x-y (solving the second equation for h). Plug that into the third equation. so: 10x+10y+20(120-x-y)=1700. Solve that for either x or y, and plug it into the first equation. It's a lot of paperwork, but keep clean notes :)

Answer 9

thank you

Answer 10

ur welcome.